Question

How do you find the amplitude, period, vertical and phase shift and graph `y=3csc[1/2(theta+60)]-3.5`?

Answer 1

See explanation and graphical depiction.

Explanation:

csc values `notin ( - 1, 1 )`. `theta` is x, here..

`y = 3 csc ( 1/2 ( x + pi/3 )) - 3.5`

`notin ( 3(-1)-3.5, 3( 1 ) - 3.5 ) = ( -6.5, - 0.5 )`

`x ne` ( asymptotic zeros of `sin (1/2 ( x + pi/3 ))`

`( 2 k + 1 ) pi/3,`

k = 0, +-1, +-2, +-3, ...#

The period is the period of `sin (1/2 ( x + pi/3 )): (2pi)/(1/2)= 4pi`

Twisted `oo`-wave amplitude : 1/2 ( period ) = `3pi`.

Vertical shift `= - 3.5 rArr` axis is`y = - 3. 5`

Phase shift = `-1/3pi`..

See graph, depicting all these aspects.
graph{(1/3(y+3.5) sin (1/2(x+1/3pi))-1)(x+7/3pi +0.0001y)(x-5/3pi +0.0001y)(y+3.5)(y+0.5) (y+6.5)(x + pi/3 +0.0001y)=0[ -10 10 -13.5 6.5]}

Here, the marked period : `x in ( - 7/3 pi, 5/3 pi )`

The phase shift marker: x = - pi/3